Difference between revisions of "Euler's totient function"

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'''Euler's totient function''', <math>\phi(n)</math>, determines the number of integers less than a given positive integer that are [[relatively prime]] to that integer.
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'''Euler's totient function''' <math>\phi(n)</math> applied to a [[positive integer]] <math>n</math> is defined to be the number of positive integers less than or equal to <math>n</math> that are [[relatively prime]] to <math>n</math>.  <math>\phi(n)</math> is read "phi of n."
  
=== Formulas ===
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==Video==
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https://youtu.be/T7INnve59JM
  
Given the [[prime factorization]] of <math>{n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_n^{e_n}</math>, then one formula for <math>\phi(n)</math> is <math> \phi(n) = n\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right) \cdots \left(1-\frac{1}{p_n}\right) </math>.
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== Formulas ==
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To derive the formula, let us first define the [[prime factorization]] of <math> n </math> as <math> n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m} </math> where the <math>p_i </math> are distinct [[prime number]]s.  Now, we can use a [[PIE]] argument to count the number of numbers less than or equal to  <math> n </math> that are relatively prime to it.
  
=== Identities ===
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First, let's count the complement of what we want (i.e. all the numbers less than or equal to <math> n </math> that share a common factor with it).  There are <math> \frac{n}{p_1} </math> positive integers less than or equal to <math> n </math> that are divisible by <math> p_1 </math>.  If we do the same for each <math> p_i </math> and add these up, we get
  
For [[prime]] p, <math>\phi(p)=p-1</math>, because all numbers less than <math>{p}</math> are relatively prime to it.
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<center><cmath> \frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.</cmath></center>
  
For relatively prime <math>{a}, {b}</math>, <math> \phi{(a)}\phi{(b)} = \phi{(ab)} </math>.
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But we are obviously overcounting.  We then subtract out those divisible by two of the <math> p_i </math>.  There are <math>\sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}</math> such numbers. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is
  
In fact, we also have for any <math>{a}, {b}</math> that <math>\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})</math>.
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<center><cmath> \sum_{1 \le i \le m}\frac{n}{p_i}
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- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}
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+ \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}. </cmath></center>
  
For any <math>n</math>, we have <math>\sum_{d|n}\phi(d)=n</math> where the sum is taken over all divisors d of <math> n </math>.
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This sum represents the number of numbers less than <math>n</math> sharing a common factor with <math>n</math>, so
  
=== See also ===
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<math>\phi(n) = n - \left(\sum_{1 \le i \le m}\frac{n}{p_i}- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}
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+ \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}\right)</math>
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 +
<math>\phi(n)= n\left(1 - \sum_{1 \le i \le m}\frac{1}{p_i}
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+ \sum_{1 \le i_1 < i_2 \le m}\frac{1}{p_{i_1}p_{i_2}} - \cdots + (-1)^{m}\frac{1}{p_1p_2\ldots p_m}\right)</math>
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<math>\phi(n)= n\left(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).</math>
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Given the general [[prime factorization]] of <math>{n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_m^{e_m}</math>, one can compute <math>\phi(n)</math> using the formula <cmath>\phi(n)= n\left(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).</cmath>
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*Note: Another way to find the closed form for <math>\phi(n)</math> is to show that the function is multiplicative, and then breaking up <math>\phi(n)</math> into its prime factorization.
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== Identities ==
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(a) For [[prime]] <math>p</math>, <math>\phi(p)=p-1</math>, because all numbers less than <math>{p}</math> are relatively prime to it.
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(b) For relatively prime <math>{a}, {b}</math>, <math> \phi{(a)}\phi{(b)} = \phi{(ab)} </math>.
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(c) In fact, we also have for any <math>{a}, {b}</math> that <math>\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})</math>.
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(d) If <math>p</math> is prime and <math>n\ge{1},</math> then <math>\phi(p^n)=p^n-p^{n-1}</math>
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(e) For any <math>n</math>, we have <math>\sum_{d|n}\phi(d)=n</math> where the sum is taken over all divisors <math>d</math> of <math>n</math>.
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Proof. Split the set <math>\{1,2,\ldots,n\}</math> into disjoint sets <math>A_d</math> where for all <math>d\mid n</math> we have
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<cmath>A_d=\{x:1\leq x\leq n\quad\text{and}\quad \operatorname{syt}(x,n)=d \}.</cmath>
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Now <math>\operatorname{gcd}(dx,n)=d</math> if and only if <math>\operatorname{gcd}(x,n/d)=1</math>. Furthermore, <math>1\leq dx\leq n</math> if and only if <math>1\leq x\leq n/d</math>. Now one can see that the number of elements of <math>A_d</math> equals the number of elements of
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<cmath>A_d^\prime=\{x:1\leq x \leq n/d\quad\text{and}\quad \operatorname{gcd}(x,n/d)=1 \}.</cmath>
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Thus by the definition of Euler's phi we have that <math>|A_d^\prime|=\phi (n/d)</math>. As every integer <math>i</math> which satisfies <math>1\leq i\leq n</math> belongs in exactly one of the sets <math>A_d</math>, we have that
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<cmath>n=\sum_{d \mid n}\phi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).</cmath>
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(f) Another interesting thing to note is that <math>\phi(n)\leq n - 1</math>. This does seem very obvious but is helpful in solving many problems.
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==Notation==
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Sometimes, instead of  <math>\phi</math>, <math>\varphi</math> is used. This variation of the Greek letter ''phi'' is common in textbooks, and is standard usage on the English [[Wikipedia]]
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== See Also ==
  
 
* [[Number theory]]
 
* [[Number theory]]
 
* [[Prime]]
 
* [[Prime]]
 
* [[Euler's Totient Theorem]]
 
* [[Euler's Totient Theorem]]
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[[Category:Functions]]
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[[Category:Number theory]]

Latest revision as of 15:30, 7 August 2020

Euler's totient function $\phi(n)$ applied to a positive integer $n$ is defined to be the number of positive integers less than or equal to $n$ that are relatively prime to $n$. $\phi(n)$ is read "phi of n."

Video

https://youtu.be/T7INnve59JM

Formulas

To derive the formula, let us first define the prime factorization of $n$ as $n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m}$ where the $p_i$ are distinct prime numbers. Now, we can use a PIE argument to count the number of numbers less than or equal to $n$ that are relatively prime to it.

First, let's count the complement of what we want (i.e. all the numbers less than or equal to $n$ that share a common factor with it). There are $\frac{n}{p_1}$ positive integers less than or equal to $n$ that are divisible by $p_1$. If we do the same for each $p_i$ and add these up, we get

\[\frac{n}{p_1} + \frac{n}{p_2} + \cdots + \frac{n}{p_m} = \sum^m_{i=1}\frac{n}{p_i}.\]

But we are obviously overcounting. We then subtract out those divisible by two of the $p_i$. There are $\sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}}$ such numbers. We continue with this PIE argument to figure out that the number of elements in the complement of what we want is

\[\sum_{1 \le i \le m}\frac{n}{p_i} - \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}.\]

This sum represents the number of numbers less than $n$ sharing a common factor with $n$, so

$\phi(n) = n - \left(\sum_{1 \le i \le m}\frac{n}{p_i}- \sum_{1 \le i_1 < i_2 \le m}\frac{n}{p_{i_1}p_{i_2}} + \cdots + (-1)^{m+1}\frac{n}{p_1p_2\ldots p_m}\right)$

$\phi(n)= n\left(1 - \sum_{1 \le i \le m}\frac{1}{p_i} + \sum_{1 \le i_1 < i_2 \le m}\frac{1}{p_{i_1}p_{i_2}} - \cdots + (-1)^{m}\frac{1}{p_1p_2\ldots p_m}\right)$

$\phi(n)= n\left(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).$

Given the general prime factorization of ${n} = {p}_1^{e_1}{p}_2^{e_2} \cdots {p}_m^{e_m}$, one can compute $\phi(n)$ using the formula \[\phi(n)= n\left(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).\]

  • Note: Another way to find the closed form for $\phi(n)$ is to show that the function is multiplicative, and then breaking up $\phi(n)$ into its prime factorization.

Identities

(a) For prime $p$, $\phi(p)=p-1$, because all numbers less than ${p}$ are relatively prime to it.

(b) For relatively prime ${a}, {b}$, $\phi{(a)}\phi{(b)} = \phi{(ab)}$.

(c) In fact, we also have for any ${a}, {b}$ that $\phi{(a)}\phi{(b)}\gcd(a,b)=\phi{(ab)}\phi({\gcd(a,b)})$.

(d) If $p$ is prime and $n\ge{1},$ then $\phi(p^n)=p^n-p^{n-1}$

(e) For any $n$, we have $\sum_{d|n}\phi(d)=n$ where the sum is taken over all divisors $d$ of $n$.

Proof. Split the set $\{1,2,\ldots,n\}$ into disjoint sets $A_d$ where for all $d\mid n$ we have \[A_d=\{x:1\leq x\leq n\quad\text{and}\quad \operatorname{syt}(x,n)=d \}.\] Now $\operatorname{gcd}(dx,n)=d$ if and only if $\operatorname{gcd}(x,n/d)=1$. Furthermore, $1\leq dx\leq n$ if and only if $1\leq x\leq n/d$. Now one can see that the number of elements of $A_d$ equals the number of elements of \[A_d^\prime=\{x:1\leq x \leq n/d\quad\text{and}\quad \operatorname{gcd}(x,n/d)=1 \}.\] Thus by the definition of Euler's phi we have that $|A_d^\prime|=\phi (n/d)$. As every integer $i$ which satisfies $1\leq i\leq n$ belongs in exactly one of the sets $A_d$, we have that \[n=\sum_{d \mid n}\phi \left (\frac{n}{d} \right )=\sum_{d \mid n}\phi (d).\]


(f) Another interesting thing to note is that $\phi(n)\leq n - 1$. This does seem very obvious but is helpful in solving many problems.

Notation

Sometimes, instead of $\phi$, $\varphi$ is used. This variation of the Greek letter phi is common in textbooks, and is standard usage on the English Wikipedia

See Also

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